Download A Relationship Between the Fibonacci Sequence and Cantor`s

UNF Digital Commons
UNF Theses and Dissertations
Student Scholarship
1994
A Relationship Between the Fibonacci Sequence
and Cantor's Ternary Set
John David Samons
University of North Florida
Suggested Citation
Samons, John David, "A Relationship Between the Fibonacci Sequence and Cantor's Ternary Set" (1994). UNF Theses and
Dissertations. Paper 285.
http://digitalcommons.unf.edu/etd/285
This Master's Thesis is brought to you for free and open access by the
Student Scholarship at UNF Digital Commons. It has been accepted for
inclusion in UNF Theses and Dissertations by an authorized administrator
of UNF Digital Commons. For more information, please contact
[email protected].
© 1994 All Rights Reserved
A Relationship Between the Fibonacci Sequence
and Cantor's Ternary Set
by
John David Samons
A thesis submitted to the Department of Mathematics and Statistics
in partial fulfillment of the requirements for the degree of
Masters of Arts in Mathematical Science
University of North Florida
College of Arts and Sciences
May, 1994
The thesis of John David Samons Is approved:
Signature Deleted
Signature Deleted
Signature Deleted
date
f)?;;;&I
11 r;.rl t4
4/ar/Ll'f
I
Signature Deleted
Chairperson
ted for the College/School:
Signature Deleted
Signature Deleted
·...
For my wife Janna for all of her support
and understanding and for having to endure the
many long nights that I worked on this thesis.
iii
ACKNOWLEDGEMENTS
My most sincere gratitude goes to Dr. Jingcheng Tong. I wish to
thank Dr. Tong for his constant encouragement throughout my graduate
program, technical guidance, academic assistance, and concern for my
graduate success. The guidance, support, and immense mathematical
knowledge that I received from Dr. Tong while I was working on my thesis
has been greatly appreciated. I will carry with me the beauty of
mathematics that Dr. Tong has bestowed on me.
I appreciate the time and effort that Dr. Scott Hochwald and Dr.
Peter Braza put into reading my thesis. The suggestions given to me by
them helped greatly in the successful completion of my thesis.
I am grateful to other members of the University of North Florida
math faculty as well. I wish to thank Dr. Champak Panchal for his
planning of my graduate program and his guidance throughout. I
appreciate the dedication to student success, the hard work, and the
availability of the mathematics instructors that I have had the privilege to
learn from.
iv
TABLE OF CONTENTS
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi
Chapter 1: Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2: The Fibonacci Sequence
Chapter 3: Cantor's Ternary Set
........................ 3
........................... 9
Chapter 4: Preliminaries . . . . . . . . . . . . . . . . . . . . . .
. . . 13
Chapter 5: The Function F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Chapter 6: The Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Chapter 7: Gaps in the Closure of F(N) . . . . . . . . . . . . . . . . . . . . . . 34
Chapter 8: Conclusion and Suggestions for Future Research . . . . . . 42
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Vita . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
v
ABSTRACT
A Relationship Between the Fibonacci Sequence
and Cantor's Ternary Set
The Fibonacci sequence and Cantor's ternary set are two objects of study
in mathematics. There is much theory published about these two objects,
individually. This paper provides a fascinating relationship between the
Fibonacci sequence and Cantor's ternary set. It is a fact that every
natural number can be expressed as the sum of distinct Fibonacci
numbers. This expression is unique if and only if no two consecutive
Fibonacci numbers are used in the expression--this is known as
Zekendorf's representation of natural numbers. By Zekendorf's
representation, a function F from the natural numbers into [0,0.603] will be
defined which has the property that the closure of F(N) is homeomorphic
to Cantor's ternary set. To accomplish this, it is shown that the closure of
F(N) is a perfect, compact, totally disconnected metric space. This then
shows that the closure of F(N) is homeomorphic to Cantor's ternary set
and thereby establishing a relationship between the Fibonacci sequence
and Cantor's ternary set.
vi
CHAPTER 1
INTRODUCTION
There are many interesting objects that are studied in mathematics.
Two such objects are the Fibonacci sequence and Cantor's ternary set.
The Fibonacci sequence is studied in such disciplines as elementary
number theory and combinatorics while Cantor's ternary set is studied in
topology and real analysis. Much theory exists concerning each object,
individually. However, in this thesis, an interesting relationship between
the Fibonacci sequence and Cantor's ternary set is established.
The Fibonacci numbers form a recursive sequence. It is a fact that
every natural number can be expressed as the sum of distinct Fibonacci
numbers. (See page 10.) However, Zekendorf found that this
representation is unique if and only if no two consecutive Fibonacci
numbers are used in the sum. This representation will prove to be an
important key to establishing the desired relationship.
The basis for the relationship is a function F that maps the natural
numbers x into [0,0.603], where x is expressed using Zekendorf's
representation. Some properties of the Fibonacci numbers will be proved
and used to establish important properties of F. It is known that Cantor's
ternary set is a perfect, compact, totally disconnected metric space. It will
be shown that for the set of natural numbers N, the closure of F(N),
denoted by ci(F), satisfies these conditions as well.
It is a fact that any two perfect, compact, totally disconnected metric
spaces are homeomorphic to each other [4]. This then implies that ci(F)
is homeomorphic to Cantor's ternary set. Two sets are said to be
homeomorphic if there exists a one-to-one function that maps one of the
sets onto the other; whereby, this one-to-one function and its inverse are
continuous. The homeomorphism between ci(F) and Cantor's ternary set
provides the relationship between the Fibonacci sequence and Cantor's
ternary set.
2
CHAPTER 2
THE FIBONACCI SEQUENCE
Background information on Fibonacci and his sequence will provide
a better understanding of and appreciation for this person and his great
mathematical achievements. Some properties of the Fibonacci sequence
are presented in order to provide a better knowledge of this powerful
sequence.
Leonardo of Pisa, better known as Leonardo Fibonacci, was one of
the most talented mathematicians of the Middle Ages. He is responsible
for many advances in the study of discrete mathematics. He was born in
about 1180 probably in Pisa, Italy. His father, Guglielmo, was appointed
chief magistrate over the community of Pisan merchants in the north
African port of Bugia (now Bejaia, Algeria). It is Leonardo's father who
helped to enhance his understanding of mathematics, for he sent
Leonardo to study calculation with an Arab master.
Fibonacci's extended trips to Egypt, Sicily, Greece, and Syria
brought him in contact with eastern and Arabic mathematical practices.
3
He was so impressed with the superiority of the Hindu-Arabic methods of
calculation that in 1202 he wrote the Liber Abaci. This book, devoted to
arithmetic and elementary algebra, illustrates and advocates Hindu-Arabic
notation and at the time helped to introduce these numerals into Europe.
Featured are the nine Indian figures--the digits 9, 8, 7, 6, 5, 4, 3, 2, and 1
and the notion that with these 9 digits and the sign 0, any number may be
written. Also included in the book are methods of calculation with
integers and fractions, computation of square roots and cube roots, and
the solution of linear and quadratic equations by false position and by
algebraic processes [2].
One of the most interesting problems posed and solved by
Fibonacci, which is found in the Liber Abaci, is the following:
A man put one pair of adult rabbits (of opposite sex) in a
certain place entirely surrounded by a wall. Assume that
each pair of adult rabbits produce one pair of young (of
opposite sex) each month. It takes two months for each pair
of young to become adults, at which time they produce their
first pair. How many pairs of rabbits are present at the
be.ginning of each month \[8]?
Assuming that none of the rabbits die, then a pair is born during the first
month, so at the beginning of the second month there are two pairs
4
present. During the second month, the original pair has produced another
pair. A month later, the original pair and the firstborn pair have produced
new pairs, so now there is a total of five pairs, and so on. The following
table illustrates the solution for the first ten months. (The numbers in the
table indicate the count at the beginning of each month.)
MONTHS
ADULT
PAl RS
1
2
3
4
5
6
7
8
9
10
1
1
1
2
3
5
8
13
21
34
1-MONTH-OLD
PAIRS
0
0
1
1
2
3
5
8
13
21
NEWBORN
PAIRS
TOTAL
PAIRS
0
1
1
2
3
5
8
13
21
34
1
2
3
5
8
13
21
34
55
89
If the first term is defined to be 1, then when the above pattern is
continued indefinitely, the sequence formed is
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,233,377,610,987, ....
The importance of the solution to this problem is that it is the Fibonacci
sequence.
5
Definition 2.1. The Fibonacci sequence, defined recursively, is
u1
= u2 = 1' un = un-2 + un-1
for natural numbers n ;:::: 3 .•
A table of the first fifteen Fibonacci numbers will serve as a useful
reference.
u1
=
1
u6
u2
= 1
u?
= 8
= 13
u3
=
2
Ua
=
21
u13 = 233
u4
=
3
Ug
=
34
u14
Us
=
5
u10 =
55
u1s = 610
U11 =
u12
89
= 144
= 377
The Fibonacci numbers form a recursive sequence. Fibonacci
solved the rabbit problem; however, he did not write down this recursive
rule for the sequence. This rule was not written until 1634 by Albert
Girard.
The Fibonacci numbers possess· a variety of remarkable properties.
One such property follows.
6
Definition 2.2. The Fibonacci sequence grows at the variable rate
Gk = uk+/uk fork = 1, 2, 3, ....111
We have
G1
= UtU1 = 1/1 =
1.000
G2
= U/U2 = 2/1 =
2.000
G3
= U/U3 = 3/2 =
1.500
G4
= usfu 4 = 5/3 =
1.667
G5
= U6/U 5 = 8/5 =
1.600
G6
= U/U6 = 13/8 =
1.625
G7
= U8/U 7 = 21/13 = 1.615
G8
= uglu 8 = 34/21 = 1.619
Theorem 2.3. The limit of Gk as k
't =
~
oo
is
't,
....
where
(1 + 5 112)/2 = 1.618034 .... This limit is called the golden ratio [8].
Proof: The recursive formula for the Fibonacci numbers
uk = uk_ 2 + uk_1 gives the characteristic equation x - x - 1 = 0. By the
quadratic formula, the characteristic roots are a = (1 + 5112)/2 and
B = (1 - 5 112)/2. Next, by induction we will show that uk = (<l- Bk)/(a - B)
for all natural numbers k.
U1 = {[(1 + 5 112)/2f - [(1 - 5112 )/2f}/{(1 + 5112)/2 - (1 - 5 112)/2} = 1.
u2 = {[(1 + 5 112 }/2] 2 - [(1 - 5112 }/2f}/{(1 + 5 112 }/2 - (1 - 5112 }/2}
= 1.
Suppose uk = (ak- Bk)/(a- B).
Now, (ak+ 1 - Bk+1)/(a- B)
= (ak- 1a 2 - Bk-1B2)/(a- B)
= [ak- 1(a
+ 1) - Bk- 1(B + 1)]/[a- B]
since a 2 - a- 1 = 0 and B2 - B - 1 = 0.
7
Therefore,
lim Gk = lim [uk+/uk]
k~oo
k~oo
=:=
lim [(ak+ 1 - Bk+ 1)/(a - B)]/[(ak - Bk)/(a - B)]
k~oo
= lim
[(ak+1 - Bk+1)/(ak- Bk)]
k~oo
= lim
{[ak+ 1/ak - Bk+ 1/ak]/[ak/ak - Bk/ak]}
k~oo
=a
si nee -1 < B < 0
This ratio has the property that if one divides the line AB at C so
that 't
= AB/AC, then
AB/AC
= AC/CB.
This number 't plays an important
role in art, for rectangles with sides in the ratio 1:1 (called golden
rectangles) are considered to be the most aesthetic. In fact, an entire
book (De Divina Proportione by Piero della Francesca) was written about
the applications of 'tin the work of Leonardo de Vinci.
8
CHAPTER 3
CANTOR'S TERNARY SET
Georg Cantor is remembered chiefly for founding set theory, one of
the greatest achievements of 19th-century mathematics. Cantor was born
in St. Petersburg, Russia, but he spent most of his life in Germany. He
took a strong interest in the arguments of medieval theologians
concerning continuity and the infinite. He studied philosophy, physics,
and mathematics in Zurich, Gottingen, and Berlin. Cantor attended the
University of Berlin, where he learned higher mathematics from Karl
Weierstrass, Ernst Kummer, and Leopold Kronecker. His doctoral thesis
was titled "In Mathematics the Art of Asking Questions is More Valuable
Than Solving Problems". He joined the faculty at the University of Halle,
first as a lecturer, then as an assistant professor, then as a full professor
in 1879.
One of his most intriguing discoveries is now known as Cantor's
ternary set (or Cantor's discontinuum or Cantor's set).
9
Definition 3.1. Cantor's ternary set C is the set of points in [0, 1]
that remain after the "middle third" intervals have been successively
removed .•
To construct Cantor's ternary set, begin by removing the middle
third of [0, 1). Let C 1 be the points that remain, so
c1
= [O, 1/3]
u [2/3, 1J.
Then remove the middle third of each of the intervals of C 1 • Let C2 be the
set that remains, so
C2 = [0,1/9)
u [2/9,1/3] u [2/3,7/9) u [8/9,1).
Removing the middle thirds again yields
C3
= [0,1/27]U
[2/3, 19/27]
[2/27,1/9]
u [2/9,7/27) u [8/27,1/3) u
u [20/27,7/9] u [8/9,25/27] u [26/27, 1).
Continuing in this manner yields C4 , C 5 ,
... ,
Cn, ....
Cantor's ternary set C is the infinite intersection of the Cn's [7].
The following property will be useful in showing that Cantor's
ternary set is totally disconnected.
10
Property 3.2. Cantor's ternary set has measure zero.
Proof: The length of the interval removed from [0, 1] to construct
C1 is 1/3. The total length of the intervals removed from [0, 1] to construct
c2 is 1/3 + 2/9, to construct c3 is 1/3 + 2/9 + 4/27,
... , to construct en is
1/3 + 2/9 + 4/27 + 8/81 + ... + 2n- 1/3n. So, the total length of the
intervals removed from [0, 1] to construct Cantor's ternary set is the infinite
sum of [2n- 1/3n]
= :En:1 (1 /2)(2/3t
This is a convergent geometric series
with sum 1. Since the total length of the removed intervals is 1 and the
length of [0, 1] is 1, Cantor's ternary set has measure zero [1 ].•
A second definition of Cantor's ternary set is presented to provide a
different view of the set. This definition offers a better understanding as
to which points of [0, 1] belong to the set.
Definition 3.3. (Alternative definition of Cantor's ternary set)
Cantor's ternary set is the set C of real numbers in [0, 1] which have a
ternary (base 3) expansion using only the digits 0 and 2 [3] .•
In other words, since every real number x in [0, 1] can be expressed
..,
in base three as x
which an
=1=
= :En=1an/3n,
where an
= 0,
1, or 2, the set of all x in
1 for all n is Cantor's ternary set [6]. So, it is easy to see that
11
1/3 is in Cantor's ternary set since
1/3
= (0.0222 ... )3
= 0/3
+ 2/9 + 2/27 + · ·· + 2/3n + ···
00
= (2/9):En=0 (1 13t = 1/3
We can find other points in the set. For example,
X =
(0.002002 ... )3
= :E:1 [2/33n]
3
9
6
= 2/3 + 2/3 + 2/3 + ...
3
3
6
= (1/3 )(2 + 2/3 + 2/3 + ···) which implies
x = (1 /3 3 )(2 + x) which implies
x = 1/13 is a point in Cantor's ternary set.
Also, we know that 20/27 is in Cantor's ternary set since
20/27
= 2/3
3
5
+ 0/3 2 + 2/3 + 0/3 4 + 0/3 + ...
= (0.202000 ... )3.
There are an infinite number of points in Cantor's ternary set. This
is seen by noting that (0.2) 3, (0.22) 3, (0.222) 3, (0.2222) 3, ... are all in the
set. In fact, Cantor's ternary set can be put into a one-to-one
correspondence with the points of [0, 1]. These are just a few of the
interesting facts about Cantor's ternary set.
12
CHAPTER 4
PRELIMINARIES
Recall Definition 2.1 (page 6) of the Fibonacci sequence.
Zekendorf found that under certain conditions that a natural number can
be written uniquely as the sum of distinct Fibonacci numbers. Before
examining Zekendorf's representation of natural numbers, we need the
following theorem.
Theorem 4.1 .. Every natural number N can be expressed as the
sum of distinct Fibonacci numbers uh < u12 < ub < ... < u1k where the u1w
(1 :::; w :::; k) are elements of the subset of Fibonacci numbers
Proof: Note that the Fibonacci numbers un,, un2• un3• ... , Um used in
this proof form a decreasing sequence. Since the Fibonacci sequence
{un}
--7 oo
as n
--7 oo,
un, :::; N < un1+ 1 for some n1 •
Notice that N - un, < un1+ 1
-
un,
= un,_ 1 < un,·
13
If N = un,, then done--otherwise,
The above shows Un 2< un,, similar reasoning shows Un3 < Un 2, and so on.
If N - un, = Un2 then done--otherwise,
Continuing in this manner results in either
N - un1 - un2 - ... - unk-1
= 1 which
implies
Zekendorf found that if no consecutive Fibonacci numbers un, un+ 1
are used in the sum, then the representation is unique. For the proof that
this representation is unique, see [9]. In summary, we have the following:
14
Property 4.2. (Zekendorf's representation of natural numbers)
For any given natural number N, there are Fibonacci numbers
conditions are satisfied:
Z1.
i1 < i2 < ... < i.
k•
Z2.
lim - inl ~ 2 for m =f n and 1 ~ m,n ~ k;
Z3.
N
= u11
+ u12 + ... + u1k and the representation is unique.
for m =f n, 1 ~ m,n ~ h, then k
= h and
im
= jm for
1 ~ m ~ k [9].e
Zekendorf's representation of 108, for example, is
108
=
1
+
5
+
13
+
89
Notice that the above three conditions are satisfied; therefore, the above
representation of 108 is unique.
Some preliminary information about Cantor's ternary set and basic
set properties are important for understanding the relationship between
the Fibonacci sequence and Cantor's ternary set. We will explore the
concept of a metric space, the idea of compactness in the set of real
numbers, the concept of a perfect set, and the idea of a totally
disconnected set.
15
Recall that a metric space is a set S together with a distance
function d which satisfies the following properties:
(1) d(x,y) ~ 0 for all points x andy in S, d(x,y) = 0 if and only if x = y,
(2) d(x,y)
= d(y,x)
for all points x and y in S, and
(3) d(x,y) + d(y,z) ~ d(x,z) for all points x, y, and z in S.
The distance function d used with Cantor's ternary set is defined by
d(x,y)
= lx-yl
for all x and y in Cantor's ternary set.
Property 4.3. Cantor's ternary set C with the distance function d is
a metric space.
Proof: The set of real numbers R with the distance function
d(x,y)
= lx-yl
is a metric space. Since Cantor's ternary set with distance
function d is a subset of R, it is a metric space.11
A subset of the real numbers with the distance function d is
compact if it is closed and bounded. This is the Heine-Borel theorem.
This proof can be found in many topology books, specifically see [3].
16
Property 4.4. Cantor's ternary set C with distance function d is
compact.
Proof: Since C is the intersection of the closed sets Cn, it is
closed. Clearly, C is a subset of [0, 1]; thus, it is bounded. Therefore, by
the Heine-Borel theorem, Cantor's ternary set with distance function d is
compact.•
Recall that a space S is perfect if every point in S is a limit point of
S. A point p is defined to be a limit point of a set S if each open disc
centered at p with radius
E
contains a point of S other than p.
Property 4.5. Cantor's ternary set C with distance function d is
perfect.
Proof: Let x be an element of C and
be a positive integer for which 2/3N <
E.
E
a positive number. Let N
Since x
= (O.x 1x2x3 ••• ) 3 in
ternary expansion where each xn is 0 or 2, we let y
C has a
= (O.y 1y2y3 ••• ) 3 be the
real number having the indicated ternary expansion:
Yn = xn for n =!= N and YN differing from xN as follows:
YN is 0 if xN is 2, and YN is 2 if xN is 0.
Then y is an element of C, and lx- yl
of C within distance
E
= 2/3N
<E. Thus, y is an element
of x, so x is a limit point of C. Since x is any point
17
of C, every point of C is a limit point of C. Therefore, Cantor's ternary set
is perfect [3]. 111
A set S is totally disconnected provided that every component of S
consists of a single point. A component S0 of a set S is a maximal
connected subset. In other words, S0 is a connected subset of S such
that there is no connected set in S containing S0 , other than S0 itself [5].
A connected set S is a set that cannot be expressed as the union of two
disjoint, non-empty open sets. We need to explore the concept of
connected sets on the real line R or on any subinterval of R. We must
show that 1) intervals on R are connected sets and 2) the only
connected sets on R are precisely the intervals.
Lemma 4.6. The real line R as well as any subinterval of R with
the usual topology is connected.
Proof: It will suffice to show this for the real line R, then it will
follow for any subinterval of R. Suppose R is disconnected. Then
R = A U B for some disjoint, non-empty open sets A and B of R. Since
A
= R\B (all points in R excluding the points of B) and B = R\A, A and B
are closed as well as open. Consider points a and b where a< band a
is in A and b is in B. Let A.
= A n [a,b]. Now A. is a closed and
18
bounded. subset of R which implies that it is compact and contains its
least upper bound g. Note that g t b since A and B are disjoint. So,
g < b. Since A contains no points of (g,b], (g,b] is a subset of B. This
implies g is in the closure of B. However, B is closed, so g is in B. Thus,
g is in both A and B. This contradicts the assumption that A and B are
disjoint. Therefore, R is connected. Following similar reasoning for a
subinterval of R represented as the union of two disjoint, non-empty open
sets will result in the same outcome [3].1111
Theorem 4.7. The connected subsets of R are precisely the
intervals.
Proof: By Lemma 4.6, every interval of R is connected. So, it
remains only to be proved that a subset D of R which is not an interval
must be disconnected. Let D be a subset of R that is not an interval.
Then there are members s and t in D and a real number w with s < w < t
for which w is not in D. Then there exists open sets U = (-oo, w) and
v=
(w, oo) satisfying the following properties:
01. s is in U n D, so U n D t { };
02. t is in V n D, so V n D t { };
03. U n V
= {},so U n V n D = {};and
04. D is a subset of U U V.
19
So, by the definition of disconnected sets, D is disconnected. Hence,
every connected subset of R must be an interval [3].11
Property 4.8. Cantor's ternary set C with distance function d-is
totally disconnected.
Proof: Since Cantor's ternary set C has measure zero, it cannot
contain any proper intervals. If C contained intervals, it would have
measure greater than zero. By Theorem 4. 7, the connected subsets of R
are precisely the intervals. Therefore, the components of C consist of
single points.liiiil
So, Cantor's ternary set C vyith distance function d defined by
d(x,y) = lx-yl for all points x and y in C is a compact, perfect, totally
disconnected metric space. These key facts about C were presented to
provide the framework for the relationship between the Fibonacci
sequence and Cantor's ternary set. It will be shown that when viewed
properly, Cantor's ternary set and the closure of the range of a function
that uses the Fibonacci numbers are homeomorphic.
Now, we give the conditions necessary for a set to be
homeomorphic to Cantor's ternary set.
20
Characterization of Cantor's ternary set [4]. A topological space
T is homeomorphic to Cantor's ternary set if and only if
T1. Tis a metric space;
T2. T is compact;
T3. T is perfect; and
T4. Tis totally disconnected.111
We have explored the Fibonacci sequence and some of its
properties. The construction of Cantor's ternary set C was presented,
and it was shown that Cantor's ternary set with distance function d
defined by d(x,y)
= lx-yl
for all points x and y in C is a perfect, compact,
totally disconnected metric space. It was shown that every natural
number can be expressed as the sum of distinct Fibonacci numbers.
Zekendorf found that this representation is unique if and only if two
consecutive Fibonacci numbers are not used. This uniqueness will prove
essential for defining the function that is the key to the relationship
between the Fibonacci sequence and Cantor's ternary set.
A function F that maps the natural numbers into [0,0.603] using
Zekendorf's representation will be defined. Several properties of F and
the Fibonacci numbers will be proved and used to show that F is an
injection from the set of natural numbers into [0,0.603]. CI(F) with
21
distance function d defined by d(x,y)
= lx-yl
for all x and y in ci(F) will be
shown to be a perfect, compact, totally disconnected metric space. It
then follows by the characterization of Cantor's ternary set that ci(F) is
homeomorphic to Cantor's ternary set.
22
CHAPTER 5
THE FUNCTION F
Now we will define the function F that maps the set N of natural
numbers into the interval [0,0.603]. The choice of 0.603 will be explained
later. This function F is based on Zekendorf's representation of natural
numbers. In the next chapter, we will prove that ci(F) is homeomorphic to
Cantor's ternary set.
Definition 5.1. Let N be the set of natural numbers, x be an
element of N, and x
= u11
+ u12 + ... + uik be Zekendorf's representation of
x. Define ttle function F by
Note that F is well-defined because Zekendorf's representation of x is
unique.liil
23
To illustrate how to find F(x) for a given natural number x, consider
x = 120. Find Zekendorf's representation of 120 by taking the Fibonacci
numbers (with indices that differ by at least two) that sum to 120.
120
F(120)
=
= 1/(2·2)
2
+
8
+
21
+
89
+ 1/(4·8) + 1/(8·21) + 1/(16·89)
= 0.2879046
The following table displays the computation of some values of F(x)
and is provided for reference.
X
1
Zekendorf's Rep.
F(x)
1
2
2
3
3
4
1+3
5
6
7
5
1+5
2+5
8
8
9
10
1+8
1/(2·1)
1/ 2·2
1/ 2·3
1/ 2·1 + 1/(4·3)
1/ 2·5
1/ 2·1 + 1/(4·5)
1/ 2·2 + 1/(4·5)
1/(2·8
1/(2·1 + 1/(4·8)
1/(2·2) + 1/(4·8)
2+8
= .500
= .250
= .167
= .583
= .100
= .550
= .300
= .063
= .531
= .281
..
Next, we need to examine the series l:m= 1 [1/(2mu 2m)]. We will show
that this series is convergent by comparing it to the convergent geometric
..
series l:n= 1 (~t Let an = (~)n and bn = (~t(1 /u 2n) for n in N. It is obvious
that (~)n(1 /u 2n) < (~t for n > 1 since 0 < 1/u 2n < 1. Therefore, by the
comparison test, this series is convergent.
24
00
Theorem 5.2. The sum c
= :Em=1[1/(2mu 2m)] = 0.6026368274
(to 10
decimal places) is an upper bound of F(N).
Proof: It was just shown that this series is convergent, so c is
finite. To see that this is an upper bound of F(N), examine the terms
1/(2u2), 1/(4u 4), 1/(8u6 ), etc .. Notice that since u1 is not used in
Zekendorf's representation, u2 is the smallest value that can occupy ui1 in
00
the first term of F(x)
= :Em=1[1 /(2mu 1J]
which implies 1/(2u 2 ) ~ 1/(2u 1,). Next,
u4 is the smallest value that can occupy u12 in the second term (since
Zekendorf's representation requires that the indices of the Fibonacci
numbers used differ by at least two), thus making 1/(4u 4)
Continuing in this manner shows that c
~
F(x)
~
~
1/(4u1J
0 for all x in N.1111
The results of the next two lemmas are necessary for proving
Lemma 5.5, which will be used to prove that F is a one-to-one function.
Lemma 5.3. Let u1 for i ; 1, 2, 3, ... be the Fibonacci sequence.
If i > j > 2, then 1/u 1 ~ 2/(3ui).
Proof: Since i > j implies i-1
~
j, we have
3u.I = 3(u.1- 1 + u.1- 2 )
25
= 3u1 + 3u1_1.
Hence, 3u 1 ;;::: 3u1 + 3u1_1.
(1)
Also, u1 = u1_1 + u1_2 ::::; 2u1_1.
(2)
Multiplying (2) by 3/2 yields (3/2)u1 ::::; 3u1_1
(3)
By (1) and (3), 3u 1;;::: 3u1 + (3/2)u1 = (9/2)u1
This results in 1/u 1::::; 2/(3u1). 8
Lemma 5.4. Let x be an element of N. If Zekendorf's
representation of x is u11 + u12 + ··· + u1k, then
F(x) < (4/3)[1 /(2u 1J]
= 2/(3u 1}
Proof: (using induction) It is easily seen that
u12 = u12-1 + u12-2 ;;::: 2u12-2 ; : : 2ul1.
If u1
m ;;:::
2m· 1u11 , then
So, F(x)
= 1/(2u 1J + 1/(2 2u1J + ··· + 1/(2ku 1J
::::; 1/(2u11 ) + 1/(23u1J + ··· + 1/(22k· 1u1J by induction hypothesis
2
< [1 /(2u 11 )](1 + 1/2 + ··· + 1/2 2k- 2 + ···)
= [1 /(2u 11 )][1 /(1-%)]
= (4/3)[1 /(2u 11 )]. Therefore,
F(x) < 2/(3u 11 ).1!
26
Lemma 5.5. Let x1 , x2 be in Nand x1 = uh + ... + u1 and
k
x2 = u11 + ... + u1h be Zekendorf's representations of x1 and x2 ,
Proof: We discuss the following three cases.
Case 1: u1m= uim form = 1, 2, ... , k. Then h ;-: : k + 1, otherwise
F(x 1 )
= F(x 2 ).
It is easily seen that F(x 2)
-
F(x 1) = 1/(2k+ 1u1kJ + ... + 1/(2hu 1h)
;-: : 1/(2hu1h)
> (0.1 )[1 /(2huiJ].
Case 2: There is a natural number 2 s; g s; k such that u1m= uim for
m = 1, 2, ... , g-1, but u19 t uJg· There are two subcases:
(i) ig > jg. Similar to Lemma 5.4, we can prove by induction that
g-1
F(x 1 ) - :Em= 1[1/(2mu 1J]
= 1/(29 U19 ) + 1/(29+1U19J + ... + 1/(2ku1J
s; 1/(2 9u1) + 1/(2 9+2u1) + ... + 1/(2 2k-9u1)
< [1/(2 9 U19)](1 + 1/22 + ... + 1/22k- 29 +·")
=. [1 /(2 9u1)][1 /(1
- 14)]
= (4/3)[1/(29u1)]
Since i9 > j9 , by Lemma 5.3, we have (4/3)[1/(2 9u1)] s; (8/9)[1/(29ui)].
g-1
So, F(x 1 )
-
:Em= 1[1 /(2mu 1J] < (8/9)[1/(29 ui)].
Therefore, F(x 2 )
-
F(x 1)
;-::::
1/(2uiJ + ... + 1/(29 Ui9) - F(x 1)
27
(4)
g-1
= 1/(2 9 u1) - {F(x 1)
:Em= 1 [1 /(2muiJ]}
-
> 1/(29 u1) - (8/9)[1/(29u1)] by (4)
= (1 /9)[1/(29u1)]
> (0.1 )[1 /(29 u1)]
~ (0.1 )[1 /(2hu1h)].
(ii) i9 < j9 . This case is impossible because consideration of
g-1
g-1
F(x 2)
-
:Em= 1[1 /(2mu 1J] in (i) instead of F(x 1)
F(x 1 )
-
F(x 2) > 0, which is a contradiction of F(x 2 ) > F(x 1 ).
-
:Em= 1[1 /(2muiJ] leads to
Case 3: u11 =f u1,. There are two subcases:
F(x 1) < (4/3)[1/(2uiJ] by Lemma 5.4
~
(8/9)[1 /(2u1J] by Lemma 5.3.
So, F(x1) < (8/9)[1 /(2u1J].
Then
F(x 2 )
-
F(x 1 )
~
(5)
1/(2u1J - F(x 1 )
> 1/(2u1J - (8/9)[1 /(2u 1,)] by (5)
= (1 /9)[1 /(2u1J]
> (0.1 )[1 /(2u1J]
~ (0.1 )[1 /(2hu1J].
28
F(x 2)
-
F(x 1) ~ F(x 2)
-
1/(2uh)
< (4/3)[1/(2uiJ]- 1/(2uh) by Lemma 5.4
~ (4/3)[1 /(3u 1J] - 1/(2u 1J by Lemma 5.3
= 4/(9u 1J - 1/(2u 1J
= -1/(18u 1J
< 0.
Hence, F(x 2 )
-
F(x 1 ) < 0. This contradicts the hypothesis that
F(x 1) < F(x 2). Therefore, this case is impossible.151
00
Recall that c
= :Lm= 1[1/(2mu 2m)1 = 0.6026368274 to
10 decimal places
is an upper bound for F(N). Next, we give an important property of the
function F.
Theorem 5.6. The function F is an injection from N into [0, c].
Proof: Suppose that there are x1 , x2 in N such that F(x 1 ) = F(x 2 ).
We will prove that this implies x1
X1 =
= x2 •
Assume that X1
+X
2•
Let
u11 + ··· + u1k and X2 = ui 1 + ... + uih be Zekendorfs representations
of x1 and x2 , respectively. Since Zekendorfs representation of a natural
29
number is unique, x1 =f x2 implies there is a natural number g such that
uig
=!=
u1gor without loss of generality h > k. Using the method in the proof
of Lemma 5.5, we know that i9 > j9 implies F(x 2 ) > F(x 1 ), i9 < j9 implies
F(x 1 ) > F(x 2 ), and h > k implies F(x 2 ) > F(x 1 ). These are contradictory to
the condition F(x 1 )
= F(x 2).liil
30
CHAPTER 6
THE MAIN RESULT
Now we prove the main result which is that ci(F) is homeomorphic
to Cantor's ternary set. It follows from the characterization of Cantor's
ternary set that it suffices to show that ci(F) with distance function d
defined by d(x,y) = lx-yl for all x and y in ci(F) is a perfect, compact,
totally disconnected metric space.
Theorem 6.1. CI(F) with distance function d is homeomorphic to
Cantor's ternary. set.
Proof: It has already been shown that ci(F) is a subset of the
interval [0, c]. (See Theorem 5.2 for explanation of c). As previously
discussed, it then follows that ci(F) with d(x,y)
= lx-yl
for x and y in ci(F) is
a metric space.
CI(F) is a bounded subset of R. Since ci(F) is the intersection of all
closed sets that contain F(N), it is closed. Since ci(F) is closed and
bounded on R, the Heine-Borel Theorem states that it is a compact set [3].
31
To show that ci(F) is a perfect set, we have to prove that every
point in F(N) is a limit point of F(N). Suppose p is an element of F(N).
Then there is an x in N such that p = F(x). If x = uit + uj2 + ... + uik is
Zekendorf's representation of x, then consider the sequence xr where
(r = 1, 2, 3, ... ) defined as follows:
It is easily seen that xr is in N and the above expression is Zekendorf's
representation of xr since ik+r+ 1
;?:
ik+2. A direct computation gives
Surely, F(xr) is an element of F(N) since xr is an element of N. Therefore,
lim F(xr) = p since 1/(2k+ 1 uik+r+ 1 ) ~ 0. Thus, p is a limit point of F(N) and
r~oo
since p is any point of F(N), every point of F(N) is a limit point of F(N).
So, ci(F) is a perfect set.
To show that ci(F) is totally disconnected; we have to prove that
each connected component of ci(F) is a single point. Assume that there
exists a connected component S of ci(F) which is not a single point.
Since S is a connected set and connected sets in the space of real
numbers are intervals by Theorem 4.7, we know that there are two real
numbers u, v such that u < v and the interval (u, v) is a subset of S. Let
32
q be an element of (u, v). Then q is a limit point of F(N), so there exists
a sequence xr in N such that F(xr) converges to q. Hence, there exists xro
in N such that F(xrJ is a point in (u, v). Using Zekendorf's representation,
Xro = ui, + Ui 2 + ··· + uih" Consider I
have F(N) n I
= { } since
= (F(xro)
- (0.1 )[1 /(2huiJ], F(xro)). We
x in N and F(x) < F(xrJ imply F(x) is not in I by
Lemma 5.5. Notice that I is an open set and that ci(F)
S
n I=
{ } and (u, v)
n I = { }. Hence,
n I = { }. This is impossible because letting
c = max{u, F(xrJ-(0.1 )(1/2huiJ}, we have (c, F(xrJ) is a subset of (u, v)
This contradiction proves that S must be a single point.111
33
n I.
CHAPTER 7
GAPS IN THE CLOSURE OF F(N)
In the construction of Cantor's ternary set C, the intervals of [0, 1]
that are removed leave behind gaps between the points of C. Likewise,
there are gaps in [0,0.603] with respect to ci(F). As was previously
shown, the gap between F(x 1 ) and F(x 2 ), where F(x 1 ) < F(x 2 ) for natural
numbers x1 = uit + uiz + ··· + uik and x2 = ui 1 + uiz + ··· + uih' is greater
than (0.1 )[1 /(2huiJ]. These gaps will be put into different classes, and
some of the gap lengths will be calculated. All gap lengths are accurate
to eight decimal places.
Definition 7.1. A class 1 gap is the interval G =(a, b), with
b = F(x) where the Zekendorf's representation of x has one term (x = ui)
and a is the closest point of ci(F) to b with a < b. Note G n ci(F) = { } .•
Before we can compute the length of the gap to the left of any point
b, where b = F(x) = 1/(2uit), we need to know how to find the closest point
34
to b, which we call a, where a < b. The method for computing this point a
for a class 1 gap will be presented as the following theorem.
Theorem 7.2. The point a closest to b
= F(x)
with x
= ui1
and a < b
when figuring a class 1 gap is defined as follows:
a = 1/(2u 11+1) + 1/(2 2ui1+ 3) + 1/(2 3 ui1+ 5 ) + ... + 1/(2nui1+( 2n- 1)) + ...
Proof: We must show t
~
a < b for all t in ci(F) where t < b.
First, we will show that a < b. We need to show that
1/(2nui1+( 2n- 1)) < 1/(3nu 1J for all natural numbers n.
Recall the results of Lemma 5.3, which is 1/ui
~
2/(3u1), where i > j > 2.
The equality holds only when u1 = 3 and u1 = 2, so this implies that
1/(ui1+1 )
= 2/(3ui1) only when ui1+ 1 = 3 and ui1 = 2. So, we will prove
inequality (1) for all other possibilities of ui1. We will prove (1) by
induction.
For n = 1, we have 1/(2uil+ 1) < 1/(3u 1}
This implies 1/(ui1+1 ) < 2/(3uiJ, which is true by Lemma 5.3.
Now, assume true for n = k:
1
Now, 1/(2k+ ui1+( 2(k+ 1)-1))
1/(2kuil+(2k- 1)) < 1/(3kui1).
= 1/(2k+1 u l+( 2k+ 1))
1
< 1/(2k+ 1 (3/2)ui1+ 2k) by Lemma 5.3
= 1/(3 ·2kui1+2k)
< 1/(3·2k(3/2)ui1+(2k- 1)) by Lemma 5.3
35
(1)
= 2/(32·2kuiH(2k-1))
< 2/(9·3ku 11 ) by induction hypothesis
= 2/(3·3k+ 1u11 )
Therefore, 1/(2nui1+( 2n. 1)) < 1/(3nu 1J for all natural numbers n.
00
00
So by (1), a= I:n=1[1/(2nu 11 +( 2n- 1))] < I:n= 1(1/(3nu 11 )].
{2)
Since the sum on the right of the inequality in (2) equals
a < 11(2u 11 ), which implies a < b.
Now, we must show that fort in ci(F), a ~ t for all t < b.
2
Recall a is defined as 1/(2u 11+ 1) + 1/(2 u11 +3) + ··· + 1/(2nui1+( 2n. 1)) + ···.
The Fibonacci number that is used in the first term of the expression for a
must be greater than u11 ; otherwise, a > b. So, to maximize the first term,
the Fibonacci number used must be the smallest one allowed (this is
u11+1). To maximize the second term, the smallest Fibonacci number
allowed by Zekendorf's representation is uh+ 3 . Following in this manner,
term by term, will maximize this sum, thus making a ~ t for all t in ci(F)
where t < b. Therefore, the point a, as defined above is the closest point
to b where a < b. 9
36
The values of b in this class are F(1) = 1/2, F(2) = 1/(2·2) = 1/4,
F(3) = 1/(2·3) = 1/6, F(5) = 1/(2·5) = 1/10, F(8), F(13), F(21 ), etc. since
Zekendorf's representation of
x requires only one term. We will
investigate the nature of these gaps by computing the lengths of some
gaps. First, the gap length to the left of 1/2, which is F(1 ), will be figured.
5
Notice a= 1/(2·2) + 1/(22 ·5) + 1/(23 ·13) + 1/(24 ·34) + 11(2 ·89) +
11(26 ·233) + 1/(27 ·610) + 1/(28 ·1597) + 1/(29 ·4181) + 1/(210 ·10946) +
1/(2 11 ·28657) + ··· = 0.31188712 (accurate to eight decimal places).
2
The sum of the remaining terms (not shown) is less than 1/(i ·2
16
)
+
1/(213 ·2 17 ) + 1/(214 ·2 18 ) + ··· = (1/2 28 )/(1- 1~) < 5·10- 9 . Thus, we have
accuracy to eight decimal places. Similar calculations have been
performed for the tails of the subsequent "gap" series. These are not
included. So, the length of the gap to the left of F(1) is
0.50000000 - 0.31188712 = 0.18811288 (accurate to eight decimal
places).
Now, we will compute the length of the gap to the left of 1/4, which
is F(2). Notice F(3) = 1/(2·3) = 1/6. Notice that a = 1/(2·3) + 1/(2 2 ·8) +
8
1/(23 ·21) + 1/(24 ·55) + 1/(2 5 ·144) + 1/(26 ·377) + 1/(27 ·987) + 1/(2 ·2584)
+ 1/(29 ·6765) + 1/(i 0 ·17711) + 1/(2 11 ·46368) + ... = 0.20527365.
Therefore, the length of the gap to the left of F(2) is
0.25000000 - 0.20527365 = 0.04472635.
37
Now, we will figure the length of the gap to the left of 1/6, which is
2
F(3). Note that F(5) = 1/(2·5) = 1/10. So, a = 1/(2·5) + 1/(2 ·13) +
3
4
6
7
5
1/(2 ·34) + 1/(2 ·89) + 1/(2 ·233) + 1/(2 ·610) + 1/(2 ·1597) +
8
9
1/(2 ·4181) + 1/(2 ·10946) + 1/(210 ·28657) + ··· = 0.12377526. This
makes the gap length to the left of 1/6 = 0.16666666 ~ 0.12377526 =
0.04289140.
Some additional gap lengths were figured, but the calculations are
not shown. The gap length to the left of 1/10, which is F(5), was
computed and found to be 0.02278603. Finally, the gap length to the left
of 1/16, which is F(8), was found to be 0.01494970.
The method for calculating a, the point closest to b where a < b, as
used in figuring a class 1 gap was proved. However, the method for
finding the closest point to b in the classes of gaps to follow has not been
proven. The method will be offered as a conjecture.
Conjecture 7.3. The point a closest to b = F(x) where a < b and
x = uil + ui2 + ··· + uik when figuring a class k gap where k ~ 2 is defined
2
1
3
as follows: a= 1/(2uil) + 1/(2 uiJ + 1/(2 ub) + ··· + 1/(2k- uikJ
+ 1/(2kuik+1) + 1/(2k+1uik+3) + 1/(2k+2uik+5) + ····li§l
38
Definition 7.4. A class 2 gap is the interval G = (a, b),
with b = F(x) where the Zekendorf's representation of x has two terms
(x = u1, + u12 ) and a is the closest point of ci(F) to b with a < b.
Note G n ci(F) = { }.EI
The values of b in this class are F(4), F(6), F(7), F(9), F(1 0), F(11 ),
F(14), etc. since 4 = uh + u12 = 1 + 3, 6 = 1 + 5, 7 = 2 + 5, 9 = 1 + 8,
10 = 2 + 8, etc.. First, we will figure the length of the gap to the left of
7/12, which is F(4). Notice that F(6) = 1/(2·1) + 1/(2 2 ·5) = 11/20. Notice
2
5
6
a= 1/(2·1) + 1/(2 ·5) + 1/(23 ·13) + 1/(24 ·34) + 1/(2 ·89) + 1/(2 ·233) +
7
8
9
0
11
1/(2 ·610) + 1/(2 ·1597) + 1/(2 ·4181) + 1/(i ·10946) + 1/(2 ·28657) +
··· = 0.56188763. So, the length of the gap to the left of 7/12 is
0.58333333 - 0.56188763 = 0.02144570.
Next, we will compute the length of the gap to the left of 11 /20,
which is F(6). So, a= 1/(2·1) + 1/(22 ·8) + 1/(23 ·21) + 1/(24 ·55) +
5
6
1/(2 ·144) + 1/(2 ·377) + 1/(2 7 ·987) + 1/(28 ·2584) + 1/(29 ·6765) +
10
1/(2 ·17711) + 1/(2 11 ·46368) + ... = 0.53860698. Thus, the gap length
to the left of F(6) is 0.55000000 - 0.53860698, which equals 0.01139302.
We will calculate another gap length, this one to the left of 3/10,
which is F(7). Here, a= 1/(2·2) + 1/(22 ·8) + 1/(23 ·21) + 1/(24 ·55) +
5
1/(2 ·144) + 1/(26 ·377) + 1/(2 7 ·987) + 1/(28 ·2584) + 1/(29 ·6765) +
39
1/(2 10 ·17711) + 1/(2 11 ·46368) + ··· = 0.28860698. So, the gap to the
left of F(7) is 0.30000000 - 0.28860698 = 0.01139302.
A few gap lengths were figured in class 2. Now, we will define a
class 3 gap and figure the lengths of the gaps to the left of some of the
gaps in this class.
Definition 7.5. A class 3 gap is the interval G = (a, b), with
b = F(x) where the Zekendorf's representation of x has three terms
(x = u11 + uj2 + ub) and a is the closest point of ci(F) to b with a < b.
Note G
n ci(F) = { }.liil
The values of b in this class are F(12), F(17), F(19), F(20), F(25),
F(27), F(28), etc. since 12 = 1 + 3 + 8, 17 = 1 + 3 + 13, 19 = 1 + 5 + 13,
etc.. Consider the gap to the left of 115/192, which is F(12). Notice that
2
3
4
5
6
a= 1/(2·1) + 1/(2 ·3) + 1/(2 ·13) + 1/(2 ·34) + 1/(2 ·89) + 1/(2 ·233) +
8
1/(2 7 ·61 0) + 1/(2 ·1597) + 1/(2 9 ·4181) + 1/(2 10 ·1 0946) + ... =
0.59522096. So, the length of the gap to the left of 115/192 is
0.59895833 - 0.59522096 = 0.00373737.
Now, F(17) = 1/(2·1) + 1/(4·3) + 1/(8·13) = 185/312. To find the
length of the gap to the left of 185/312, use a = 1/(2·1) + 1/(2 2 ·3) +
8
1/(23 ·21) + 1/(24 ·55) + 1/(2 5 ·144) + 1/(26 ·377) + 1/(27 ·987) + 1/(2 ·2584)
40
9
+ 1/(2 -6765) + 1/(210 ·17711) + 1/(211 ·46368) + ...
= 0.59069031.
Thus, the gap length to the left of 185/312 is 0.59294871 - 0.59069031 =
0.00225840.
The first three classes of gaps were defined. Now, we define a
class in general i.e. the n'h class.
Definition 7.6. A class n gap is the interval G = (a, b), with
b = F(x) where the Zekendorf's representation of x has n terms
(x = uil + uj2 + ... + uiJ and a is the closest point of ci(F) to b with a < b.
Note G n ci(F)
= { }.Iii
Clearly, there are an infinite number of classes each with an infinite
number of gaps. Some of the gaps of the first three classes were
computed. More computation might reveal a pattern for the gap lengths
within a specific class.
41
CHAPTER 8
CONCLUSION AND SUGGESTIONS FOR FUTURE RESEARCH
The relationship between the Fibonacci sequence and Cantor's
ternary set was found using the function F which utilizes Zekendorf's
representation of natural numbers. CI(F) with distance function d defined
by d(x,y)
= lx-yl
for all x and y in ci(F) was shown to be homeomorphic to
Cantor's ternary set. This then means that there exists a one-to-one
function H(x) from ci(F) onto Cantor's ternary set, where H(x) and H- 1(x)
are both continuous. Explicitly stating H seems to be a difficult task.
Several functions were examined before choosing the function F
defined by F(x)
= 1/(2u 11 ) + 1/(2 2uiJ + ... + 1/(2nuiJ
The function F was
shown to be one-to-one, and ci(F) was shown to be homeomorphic to
Cantor's ternary set. The function G defined by
G(x) = 1/u 11 + 1/ui2 + 1/ub + ... + 1/ui" was tried first; however, there was
some difficulty in showing that ci(G) was totally disconnected. It would be
interesting to see if ci(G) could be proved to be a perfect, compact, totally
disconnected metric space.
42
It was found that there are gaps to the left of the points of ci(F).
These gaps were put into different classes. Further computing might
uncover a pattern for the lengths of gaps within specific classes, and it
would be interesting to see if Conjecture 7.3 could be proved.
The Fibonacci sequence and Cantor's ternary set are interesting
objects studied in mathematics. Separately, much theory can be found
about the properties of each object. However, this thesis has presented
theory that establishes an intriguing relationship between two seemingly
different areas of study. A relationship between the Fibonacci sequence
and Cantor's ternary set was established. Continued research into this
area is encouraged to determine if there is more to this relationship.
43
BIBLIOGRAPHY
1.
Bartle, Robert G. The Elements of Real Analysis. 2nd ed. New
York: John Wiley and Sons, 1964.
2.
Boyer, Carl B. A History of Mathematics. Princeton: Princeton
University Press, 1985.
3.
Croom, Fred H. Principles of Topology. Philadelphia: Saunders
College, 1989.
4.
Hocking, John G., and Gail S. Young. Topology. Reading: AddisonWesley, 1961.
5.
Kasriel, Robert H. Undergraduate Topology. Philadelphia: W. B.
Saunders, 1971.
6.
Kuratowski, Kazimierz. Introduction to Set Theory and Topology.
Oxford: Pergamon Press, 1972.
7.
Marsden, Jerrold E. Elementary Classical Analysis. New York: W.
H. Freeman, 197 4.
8.
Roberts, Fred S. Applied Combinatorics. Englewood Cliffs:
Princeton-Hall, 1984.
9.
Vajda, S. Fibonacci & Lucas Numbers, and the Golden Section:
Theory and Applications. Chichester: Ellis Horwood Limited,
1989.
44
Name of Author:
John David Samons
Place of Birth:
Somerville, New Jersey
Date of Birth:
EDUCATION
M.S. in Mathematics, 1994.
University of North Florida, Jacksonville, Florida.
B.S. in Mathematics Education, 1981; A.A., 1979.
University of Florida, Gainesville, Florida.
TEACHING EXPERIENCE
Adjunct Mathematics Instructor. Florida Community College at
Jacksonville, South Campus.
Advisor I I Mathematics Tutor. Learning Assistance Center,
Florida Community College at Jacksonville, Jacksonville, Florida.
Mathematics Instructor. Jacksonville Job Corps Center, Bradford
High School, Lake Butler High School.
PROFESSIONAL EXPERIENCE
Presenter. "Mathematics CLAST Remediation: Issues and
Answers." National Conference on College Teaching and Learning.
Jacksonville, Florida.
Developer. CLAST remediation program for mathematics. Florida
Community College at Jacksonville, Jacksonville, Florida.
45